Answer
$\left\{\frac{\pi}{3}+2\pi(n), \frac{5\pi}{3}+2\pi(n), \pi+2\pi(n), \text{where n is an integer}\right\}$
Work Step by Step
RECALL:
$\cos{(2x)}=2\cos^2{x}-1$
Thus, the given equation can be written as:
$\cos{(2x)}+\cos{x}=0
\\2\cos^2{x}-1+\cos{x}=0
\\2\cos^2{x}+\cos{x}-1=0$
Let $u=\cos{x}$
Replace $\cos{x}$ by $u$ to obtain:
$2u^2+u-1=0$
Factor the trinomial to obtain:
$(2u-1)(u+1)=0$
Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain:
\begin{array}{ccc}
&2u-1=0 &\text{or} &u+1=0
\\&2u=1 &\text{or} &u=-1
\\&u=\frac{1}{2} &\text{or} &u=-1
\end{array}
Replace $u$ by $\cos{x}$ to obtain:
\begin{array}{ccc}
&\cos{x}=\frac{1}{2} &\text{or} &\cos{x}=-1
\\&x=\cos^{-1}{(\frac{1}{2})} &\text{or} &x=\cos^{-1}{(-1)}
\end{array}
Use the inverse cosine function of a scientific calculator to obtain:
\begin{array}{ccc}
&x=\frac{\pi}{3}, \frac{5\pi}{3} &\text{or} &x=\pi
\end{array}
Since the period of $y=\cos{x}$ is $2\pi$, then the solutions to the given equation are the values that satisfy:
$\left\{\frac{\pi}{3}+2\pi(n), \frac{5\pi}{3}+2\pi(n), \pi+2\pi(n), \text{where n is an integer}\right\}$
